The number of transpositions in $S_n$ divides the cardinal of the conjugacy class of any odd permutation I was wondering if there is an elegant (or not so elegant) proof of the following:

There is an $n_0$ (possibly equal to one) such that for any $n\geq n_0$ the number of transpositions in $S_n$ divides the cardinal of the conjugacy class of any odd permutation.

Edit: The context of this is not very relevant. In any case, I want to prove that there is a positive linear combination of any irreducible character of $S_n$ evaluated at the odd conjugacy classes that equals zero (different from the trivial and signature characters). I know how to prove this but I would like to divide by the cardinal of the conjugacy class of a transposition to have as coefficient one in that term, the problem is that I would like to keep the coefficients in $\mathbb{N}$.
Thanks!
 A: Here is an attempt at a proof (for all values of $n$), but it is not very elegant I am afraid! Any improvements would be welcome.
Since the number of transpositions is $n(n-1)/2$, we need to trove that the $|C_G(\sigma)|$ divides $2(n-2)!$ for an odd permutation $\sigma \in S_n$.
Odd permutations have an odd number of cycles of even length, so there exists an even $r$ and odd $t$ such that $\sigma$ has exactly $t$ cycles of length $r$. So we can write $\sigma=\sigma_1\sigma_2$, where $\sigma_1$ consists of the cycles of length $r$, and $\sigma_2$ the remaining cycles.
Then $C_G(\sigma) = C_{S_{tr}}(\sigma_1) \times C_{S_{n-tr}}(\sigma_2)$, so it is sufficient to prove the result in the case $\sigma = \sigma_1$.
So $n=tr$ and $\sigma$ consists of $t$ cycles of length $r$, and its centralizer $C_G(\sigma)$ is a wreath product $ C_{r} \wr S_t$, which has order $r^tt!$.
Now if $r>4$, then $S_{r-2}$ has a subgroup $H$ of order $r$, and so $S_{t(r-2)}$ has a subgroup $H \wr S_t$ of order $r^tt!$, so this divides $(r-2)t! \le (rt-2)!= (n-2)!$ and we are done.
So we just have to handle the cases $r=2$ and $r=4$.
Write $r^tt! = 2^km$ with $m$ odd. Then $m$ divides $t!$ and $t \le n-2$, so it remains to prove that $2^k$ divides $2(n-2)!$.
Now $2^k$ = $r^t2^j$ where $2^j$ divides $(t-1)!$.
When $r=2$, since $C_r \wr S_{t-1} \le S_{n-2}$, we have $r^{t-1}2^j$ divides $(n-2)!$, so $r^t2^j$ divides $2(n-2)!$ as claimed.
When $r=4$, we have $C_r \wr S_{t-1} \le S_{n-4}$, so $r^{t-1}2^j$ divides $(n-4)!$, and again $r^t2^j$ divides $2(n-2)!$.
